Chapter 7 Power series methods 7.1 Power series Note: 1 or 1.5 lecture , §3.1 in [EP], §5.1 in [BD] Many functions can be written in terms of a power series 1 X k ak(x − x0) : k=0 If we assume that a solution of a differential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coefficients. That is, we will try to solve for the numbers ak. Before we can carry out this process, let us review some results and concepts about power series. 7.1.1 Definition As we said, a power series is an expression such as 1 X k 2 3 ak(x − x0) = a0 + a1(x − x0) + a2(x − x0) + a3(x − x0) + ··· ; (7.1) k=0 where a0; a1; a2;:::; ak;::: and x0 are constants. Let n X k 2 3 n S n(x) = ak(x − x0) = a0 + a1(x − x0) + a2(x − x0) + a3(x − x0) + ··· + an(x − x0) ; k=0 denote the so-called partial sum. If for some x, the limit n X k lim S n(x) = lim ak(x − x0) n!1 n!1 k=0 247 248 CHAPTER 7. POWER SERIES METHODS exists, then we say that the series (7.1) converges at x. Note that for x = x0, the series always converges to a0. When (7.1) converges at any other point x , x0, we say that (7.1) is a convergent power series. In this case we write 1 n X k X k ak(x − x0) = lim ak(x − x0) : n!1 k=0 k=0 If the series does not converge for any point x , x0, we say that the series is divergent. Example 7.1.1: The series X1 1 x2 x3 xk = 1 + x + + + ··· k! 2 6 k=0 is convergent for any x. Recall that k! = 1 · 2 · 3 ··· k is the factorial. By convention we define 0! = 1. In fact, you may recall that this series converges to ex. We say that (7.1) converges absolutely at x whenever the limit n X k lim jakj jx − x0j n!1 k=0 P1 k exists. That is, if the series k=0jakj jx − x0j is convergent. Note that if (7.1) converges absolutely at x, then it converges at x. However, the opposite is not true. Example 7.1.2: The series X1 1 xk k k=1 P1 (−1)k converges absolutely at any x 2 (−1; 1). It converges at x = −1, as k=1 k converges (condition- ally) by the alternating series test. But the power series does not converge absolutely at x = −1, P1 1 because k=1 k does not converge. The series diverges at x = 1. 7.1.2 Radius of convergence If a series converges absolutely at some x1, then for all x such that jx − x0j ≤ jx0 − x1j we have k k k that jak(x − x0) j ≤ jak(x1 − x0) j for all k. As the numbers jak(x1 − x0) j sum to some finite limit, k summing smaller positive numbers jak(x − x0) j must also have a finite limit. Therefore, the series must converge absolutely at x. We have the following result. Theorem 7.1.1. For a power series (7.1), there exists a number ρ (we allow ρ = 1) called the radius of convergence such that the series converges absolutely on the interval (x0 − ρ, x0 + ρ) and diverges for x < x0 − ρ and x > x0 + ρ. We write ρ = 1 if the series converges for all x. 7.1. POWER SERIES 249 diverges converges absolutely diverges x0 − ρ x0 x0 + ρ Figure 7.1: Convergence of a power series. See Figure 7.1. In Example 7.1.1 the radius of convergence is ρ = 1 as the series converges everywhere. In Example 7.1.2 the radius of convergence is ρ = 1. We note that ρ = 0 is another way of saying that the series is divergent. A useful test for convergence of a series is the ratio test. Suppose that X1 ck k=0 is a series such that the limit ck+1 L = lim n!1 ck exists. Then the series converges absolutely if L < 1 and diverges if L > 1. k Let us apply this test to the series (7.1). That is we let ck = ak(x − x0) in the test. We let k+1 ck+1 ak+1(x − x0) ak+1 L = lim = lim = lim jx − x0j: n!1 n!1 k n!1 ck ak(x − x0) ak A Define by ak+1 A = lim : n!1 ak Then if 1 > L = Ajx − x0j the series (7.1) converges absolutely. If A = 0, then the series always 1 1 converges. If A > 0, then the series converges absolutely if jx− x0j < =A, and diverges if jx− x0j > =A. That is, the radius of convergence is 1=A. Let us summarize. Theorem 7.1.2. Let 1 X k ak(x − x0) k=0 be a power series such that ak+1 A = lim n!1 ak exists. If A = 0, then the radius of convergence of the series is 1. Otherwise the radius of convergence is 1=A. 250 CHAPTER 7. POWER SERIES METHODS Example 7.1.3: Suppose we have the series X1 2−k(x − 1)k: k=0 First we compute, −k−1 ak+1 2 −1 A = lim = lim = 2 = 1=2: −k k!1 ak k!1 2 Therefore the radius of convergence is 2, and the series converges absolutely on the interval (−1; 3). The ratio test does not always apply. That is the limit of ak+1 might not exist. There exist more ak sophisticated ways of finding the radius of convergence, but those would be beyond the scope of this chapter. 7.1.3 Analytic functions Functions represented by series are called analytic functions. Not every function is analytic, although the majority of the functions you have seen in calculus are. ∗ An analytic function f (x) is equal to its Taylor series near a point x0. That is, for x near x0 we have X1 f (k)(x ) f (x) = 0 (x − x )k; (7.2) k! 0 k=0 (k) th where f (x0) denotes the k derivative of f (x) at the point x0. -10 -5 0 5 10 3 3 2 2 1 1 0 0 -1 -1 -2 -2 -3 -3 -10 -5 0 5 10 th th Figure 7.2: The sine function and its Taylor approximations around x0 = 0 of 5 and 9 degree. ∗Named after the English mathematician Sir Brook Taylor (1685 – 1731). 7.1. POWER SERIES 251 For example, sine is an analytic function and its Taylor series around x0 = 0 is given by X1 (−1)n sin(x) = x2n+1: (2n + 1)! n=0 In Figure 7.2 on the facing page we plot sin(x) and the truncations of the series up to degree 5 and 9. You can see that the approximation is very good for x near 0, but gets worse for larger x. This is what will happen in general. To get good approximation far away from x0 you will need to take more and more terms of the Taylor series. 7.1.4 Manipulating power series One of the main properties of power series that we will use is that we can differentiate them term by P k term. That is Suppose that ak(x − x0) is a convergent power series. Then for x in the radius of convergence we have 2 1 3 1 d 6X 7 X 6 a (x − x )k7 = ka (x − x )k−1: dx 46 k 0 57 k 0 k=0 k=1 Notice that the term corresponding to k = 0 disappeared as it was constant. The radius of conver- gence of the differentiated series is the same as that of the original. Example 7.1.4: Let us show that the exponential y = ex solves y0 = y. First write X1 1 y = ex = xk: k! k=0 Now differentiate X1 1 X1 1 y0 = k xk−1 = xk−1: k! (k − 1)! k=1 k=1 For convenience we reindex the series by simply replacing k with k + 1. The series does not change, what changes is simply how we write it. After reindexing the series starts at k = 0 again. X1 1 X1 1 xk−1 = xk: (k − 1)! k! k=1 k=0 x d x x That was precisely the power series for e that we started with, so we showed that dx e = e . Convergent power series can be added and multiplied together, and multiplied by constants using the following rules. Firstly, we can add series by adding term by term, 0 1 1 0 1 1 1 BX C BX C X B − kC B − kC − k @B ak(x x0) AC + @B bk(x x0) AC = (ak + bk)(x x0) : k=0 k=0 k=0 252 CHAPTER 7. POWER SERIES METHODS We can multiply by constants, 0 1 1 1 BX C X B − kC − k α @B ak(x x0) AC = αak(x x0) : k=0 k=0 We can also multiply series together, 0 1 1 0 1 1 1 BX C BX C X B − kC B − kC − k @B ak(x x0) AC @B bk(x x0) AC = ck(x x0) ; k=0 k=0 k=0 where ck = a0bk + a1bk−1 + ··· + akb0. The radius of convergence of the sum or the product is at least the minimum of the radii of convergence of the two series involved.